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Rolle's Theorem

  1. Mar 30, 2009 #1
    1. The problem statement, all variables and given/known data
    determine whether rolle's theorem can be applied to f on the closed interval [a,b]. If Rolle's theorem can be applied, find all values of c in the open interval (a,b) such that f'(c) = 0


    2. Relevant equations
    f(x) = (x-1)(x-2)(x-3) with the interval being [1,3]

    3. The attempt at a solution
    the function is continuous on [1,3] and differentiable on (1,3)

    f(1) = 0 and f(3) = 0 so f(1) = f(3)

    f'(x) = 3x^2 - 12x + 11

    now I need to find the point in [1,3] with slope 0 so I set f'(x) = 0.

    I know from here i'm supposed to get c values of (6 - sqrt 3) / 3 and (6 + sqrt 3) / 3 however the algebra to get me to this point is eluding me.

    Can someone please point of the obvious for me?

    Thanks
     
  2. jcsd
  3. Mar 30, 2009 #2

    CompuChip

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    Science Advisor
    Homework Helper

    It is simply a quadratic equation:
    [tex]3x^2 - 12x + 11 = 0[/tex]
    You can solve it using your standard toolbox, e.g. by applying the quadrature formula:
    [tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex].
     
  4. Mar 30, 2009 #3
    use use the Quadratic equation to solve for x



    x = (-b +/- sqrt(b^2 - 4ac)) / 2a


    your equation is f'(x) 3x^2 - 12x + 11

    Set for zero

    3x^2 - 12x + 11=0


    So.

    a = 3 b = -12 c= 11

    x = (12+/- Sqrt(144 - (4*3*11)))/6

    x = (12 +/- Sqrt(12))/6

    x = (Sqrt(3)+6)/3 or -(Sqrt(3)-6)/3
     
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